Question

我写了一个小方法来查看字符串是否包含另一个字符串。我只有一个小问题，它总是返回false。

鉴于haystack是一个名为salaryCheck的字符串，其值为＆＃34; 10.1＆＃34;，并且针是＆＃34;。＆＃34;。它总是返回false。

根据我的理解，它应该返回true，我首先将所有内容放入字符向量中以便于阅读。然后我进入一个循环，在其中我检查第一个针符号是否与haystack [i]匹配。如果第一个针头符号与haystack [i]匹配，我继续进入另一个循环，我将从haystack [i]开始的所有草堆与针头符号的完整列表进行比较。

据我所知，这应该与我给出的论点一致。

这是我的代码：

bool contains(std::string& haystack, std::string needle){
    if (haystack.size() < needle.size())
        return false;

    bool returnValue = false;
    std::vector<char> haystackChars;
    std::vector<char> needleChars;

    for (char c : haystack)
        haystackChars.push_back(c);
    for (char c : needle)
        needleChars.push_back(c);

    for (int i = 0; i < haystackChars.size() && !returnValue; i++){
        if (needleChars[0] == haystackChars[i])
            for (int i2 = i; i2 < needleChars.size(); i2++){
                if (needleChars[i2] == haystackChars[i2])
                    returnValue = true;
                else{
                    returnValue = false;
                    break;
                }
            }
    }

    return returnValue;
}

Answer 1

问题出在这里

        for (int i2 = i; i2 < needleChars.size(); i2++){

您应该在0和needleChars.size()之间循环，或者i和i + needleChars.size()。以下if声明也需要调整; needle和haystack的正确数组索引不会相同。

Answer 2

几分钱，随你随心所欲。

bool contains(std::string& haystack, std::string needle){
// you only need to inspect haystack and needle, so 
bool contains(const std::string& haystack, const std::string& needle)
// is preferred. You also get convinient constructors, simplifying the 
// function call. And of course the const guarantee

// This is ok. 
    if (haystack.size() < needle.size())
        return false;
// returnValue will be unneccesary, you can leave as soon as you find the positive.
    bool returnValue = false;
// The vector does not add anything useful. std::string works fine with indexes.
    std::vector<char> haystackChars;
    std::vector<char> needleChars;

    for (char c : haystack)
        haystackChars.push_back(c);
    for (char c : needle)
        needleChars.push_back(c);

    // The algorithm is unnecessarily complex, 
    // use the string members to simplify.
    for (int i = 0; i < haystackChars.size() && !returnValue; i++){
        if (needleChars[0] == haystackChars[i])
            // This for statment can be rewritten as
            if ( haystack.substring( i, needleSize ) == needle )
                return true;
            for (int i2 = i; i2 < needleChars.size(); i2++){
                if (needleChars[i2] == haystackChars[i2])
                    returnValue = true;
                else{
                    returnValue = false;
                    break;
                }
            }
    }
    // this becomes return false. 
    return returnValue;
}

通过这些修订：

bool contains(const std::string& haystack, const std::string& needle){
    if (haystack.size() < needle.size())
        return false;
    size_t needleSize = needle.size();
    for (int i = 0; i < haystack.size(); i++){
        if ( haystack.substr( i, needleSize ) == needle )
                return true;
    }
    return false;
}

正如dyp所说，substr可能很昂贵，string ::找到救援。条件也可以写为

        if ( haystack.find( needle.c_str(), i, needleSize ) == i )
                return true;

当然，正如所建议的，如果这不是一个练习，那么这将是

if ( haystack.find( needle ) != string::npos )

C ++ - Haystack / Needle字符串检查总是返回false

2 个答案: